For starter, recall that an arithmetic progression [tex]\{a_n\}_{n\ge1}[/tex] is given recursively by
[tex]\begin{cases}a_1=a\\a_n=a_{n-1}+d&\text{for }n>1\end{cases}[/tex]
We can solve this explicitly for [tex]a_n[/tex] in terms of [tex]a_1[/tex]:
[tex]a_n=a_{n-1}+d[/tex]
[tex]a_n=(a_{n-2}+d)+d=a_{n-2}+2d[/tex]
[tex]a_n=(a_{n-3}+d)+d=a_{n-3}+3d[/tex]
and so on, down to
[tex]a_n=a_1+(n-1)d[/tex]
More generally, we can express [tex]a_n[/tex] in terms of an arbitrary term [tex]a_k[/tex]:
[tex]a_n=a_k+(n-k)d[/tex]
(notice how the index of [tex]a_k[/tex] and the coefficient of [tex]d[/tex] on the right side add to [tex]n[/tex])
Let [tex]a_n[/tex] be the [tex]n[/tex]-th term of this particular progression. We're given
[tex]\begin{cases}a_3+a_7=6\\a_3a_7=8\end{cases}[/tex]
from which it follows that
[tex]a_7=\dfrac8{a_3}\implies a_3+\dfrac8{a_3}=6[/tex]
[tex]\implies{a_3}^2-6a_3+8=(a_3-4)(a_3-2)=0\implies a_3=4\text{ or }a_3=2[/tex]
[tex]\implies a_7=2\text{ or }a_7=4[/tex]
If [tex]a_3=4[/tex], it follows that
[tex]a_7=a_3+(7-3)d\implies2=4+4d\implies d=-\dfrac12[/tex]
Otherwise, if [tex]a_3=2[/tex], then
[tex]4=2+(7-3)d\implies2=4d\implies d=\dfrac12[/tex]
The sum of the first [tex]k[/tex] terms of an arithmetic progression [tex]\{a_n\}_{n\ge1}[/tex] is
[tex]S_k=\displaystyle\sum_{n=1}^ka_n=\sum_{n=1}^k(a_1+(n-1)d)=ka_1+d\frac{k(k-1)}2[/tex]
If [tex]a_3=4[/tex], then
[tex]a_1=a_3+(1-3)d\implies a_1=5[/tex]
so that
[tex]S_{16}=16\cdot5-\dfrac12\dfrac{16\cdot15}2=20[/tex]
Otherwise, if [tex]a_3=2[/tex], then
[tex]a_1=a_3+(3-1)d\implies a_1=1[/tex]
so that
[tex]S_{16}=16\cdot1+\dfrac12\dfrac{16\cdot15}2=76[/tex]
